Category: Geometry

Gangbuster Geometry

Geometry, trigonometry, and calculus are required for this challenge.

Diagram of pond after being hit by rock. waves are just for illustration purposes and are not part of the problem.

We left an unsolved puzzle at the end of Fascinating Frequencies 3. We want to find the time it takes for a stone to stop after hitting water. The stone has an initial velocity of 1 m/s, a mass of 1 kilogram, and radius 10cm. The force acting on it from gravity is 1kg*g down. The buoyant force acting on it is the weight of the water displaced by the stone. The water has density 0.9982 kg/m3. The friction from the water in any direction acts perpendicular to the surface, inwards, with a strength equal to the sine of the angle multiplied by the density of the water.

\vec F_G=1kg g\ down\\[8pt] \vec F_B=\frac{998.2kg gV_w}{m^3}\ up\\[8pt] \vec F_d=\frac{469.2kgA_f\vec V^2}{m^3}\ up
solution

Sophisticated Shapes

In this challenge, you will need to use geometry, calculus, algebra, and angle theorems.

A roulette is a shape created by rolling things. If you roll circles, you have a trochoid. If you draw the path of a point on the circumference, you have a cycloid. And if you roll that circle inside another circle, you have a hypocycloid. Simple enough. Say you have a circle with radius 25 centimetres and a circle with radius 1 metre. When you rotate the smaller circle inside the bigger one, the path traveled by a point on the circumference is a hypocycloid. It will touch the bigger circle 4 times every rotation. At these points, the instantaneous velocity of the point is zero. The challenge is to find the instantaneous acceleration if the small circle completes 1 revolution inside the bigger one each second. This is different from rolling completely over the inside, which it will do every 4.

solution

Peculiar Products

Trigonometry, geometry, ad algebra are required to solve todays challenge.

Pi notation is to Sigma notation as plus is to minus. Sigma notation is the sum of a bunch of stuff, Pi is the product. Pi is also ≈3.1416, but we will note that as π. In Strange Sums, we evaluated an expression with 2 Sigmas. Here will will find the product of the outputs of an equation. Specifically, the equation for the area of a regular polygon. Better go find that one out right now if you don’t know it. We will express this function as A(n, s), where n is the number of sides and s the length of them. Now solve the expression below.

\prod_{k=1}^\infin\Bigg(A\bigg(k,\,\frac1{tan (90\degree)-\frac{360\degree}k}\bigg)\Bigg)
Solution

Vile Voting

Algebra and combinatorics are needed to solve this challenge.

Brutus and Craw are voting over their preference of colours. Their satisfaction is measured by how close their preferences are to the ones agreed on; they do not care about each other. The vote is determined by each player prioritizing the colours of the rainbow. Brutus and Craw do not know about indigo. Their first pick gets 5 points, their second 4, and so on such that their last gets 0.The colour with the most is first, the second most, second, and so on. Brutus and craw receive “points” of satisfaction, based on how close the vote is to their personal preferences. For each preference the number of points starts at one less than the previous, with 5 for 1st. It goes down by 2 if it’s below it’s position, and then decreases by one until it stops at 0. Except for fifth position, which cannot drop 2 because it started at only 1. In case of a tie, the position that comes first is picked randomly. First, Brutus states his preferences: red, orange, yellow, purple, blue, green. Then Craw would states his: purple, blue, green, yellow, orange, red. Except that Craw is a meanie and adjusts his to make it unfair. But Brutus then adjusts his, and they go back and forth until neither can improve their score. The challenge is to figure out what the votes, standings, and satisfaction will be for Brutus and Craw will be when they finish, if they ever do.

Satisfaction\\[16pt] \begin{matrix} & 1st & 2nd & 3rd & 4th & 5th & 6th\\ \#1 & 5 & 3 & 2 & 1 & 0 & 0\\ \#2 & 4 & 4 & 2 & 1 & 0 & 0\\ \#3 & 3 & 3 & 3 & 1 & 0 & 0\\ \#4 & 2 & 2 & 2 & 2 & 0 & 0\\ \#5 & 1 & 1 & 1 & 1 & 1 & 0\\ \#6 & 0 & 0 & 0 & 0 & 0 & 0\\ \end{matrix}
Solution

Recursive Renewals

You will need to do algebra, combinatorics, calculus, and trigonometry, and also use multiple matrices, to solve today’s challenge.

When you take a book out at the Leibniz Library, you must return it before the fine equation, f(x), rises above zero. If not, your fine is determined by the fine equation. The variable r represents the number of times you have renewed the book, beggining at zero. The variable d counts days and increases by 1 every day, beggining at zero. The variable t counts days and increases by 1 each day, beggining at 0. Each time you renew the book, the variable r in the fine equation, which begins at zero, increases by 1, and the variable t is reset to 0. You may not renew the book if someone else has it on hold or the fine equation is above 0. The probability that someone puts your book on hold on any given day is determined by the hold equation, h(x). How can you maximize the number of days you have with your book without paying anything?

f(x)=d^2+5d-7cos(r)+r^3-100\\[10pt] h(x)=\frac{t^2+3sin(t)+15}{100}
Solution

Fascinating Fruits

Diagram of the Fascinating Fruits.

You will need an understanding of algebra and combinatorics to solve today’s challenge.

Today’s challenge begins with 3 rows of fruits. They begin with an apple, and orange, and a pear. The fruits behind the front are unknown. Directly behind any apple, there is an equal chance of a plum, a banana, or an orange. There is no other fruit directly behind an apple, but one apple is at the end of a row. Behind any plum is the same fruit that was in front of it. Directly behind an orange, there is a one third probability there are grapes. There is an equal probality of pears and blueberries, and 1 orange leads to a banana. For any banana there is an equal chance of blueberries or a plum. 1 banana is followed by grapes, and another ends a row. Behind blueberries, there is a one third chance of a pear, otherwise grapes. Directly behind grapes, there is a 50% chance of more grapes, a 25% chance of a plum, and otherwise a pear. Finally, directly behind a pear there is the fruit of the colour in the rainbow following that of the fruit in front of the pair. The exception is the 1 pear that ends a row. The apples in this question are red, which is considered to follow purple, the colour of the plums. The fruit following the pair at the front is an apple. The challenge is to use this information to find the probability that the row started by a pear ends with one.

Solution

Crazy Capture

All of your algebra, geometry, trigonometry, angle theorems, and pathmaking skills are required to have a chance of solving today’s challenge.

2D capture example.

In a mathematical capture, all police officers and the target are points that travel at the same constant speed in infinite space in n dimensions. They have infinite endurance and perfect reaction time. The target is instantly captured and cannot ecape when a police officer touches it. In 1 dimension, when n = 1, it obviously takes 2 officers surrounding the target. When n = 2, we can do it with three easily. We do this by dividing the plane into three, each officer can easily trap the target in a 120° slice. The challenge is to see how many officers it takes to capture the target when n equals 3, in three dimensions.

Solution

Devising Deactivation

To solve this challenge, you will need to use geometry, trigonometry, and algebra, to do pathmaking.

Map of the radar that must be deactivated.

You are working for the government as a mathematician and they tell you of a problem. There is a plan afoot to compromise an enemy radar station. It is critical that the invading crew member is undedected, as the enemy must not know that their radar is sending them misinformation. However, the radar has a range of 100 metres and completes a rotation every 12 seconds. The fastest crew member can only sprint at 7 metres per second with their toolkit. Because of the radar’s length and rotation speed, it is nearly impossible to disable it undetected. None of the simple strategies work. The challenge is to find a path that the invader can take without being detected.

Solution

Crazy Quadrilateral

You will need to use geometry, angle theorems, trigonometry, and your brain to solve this challenge.

Crazy Quadrilateral

We begin with a quadrilateral AEGH. We then draw a line from A to G to make ΔAEG and ΔAGH. ΔAEG is an isoscelese triangle. Next, we draw a line from point E to a new point, F, such that ∠AEF measures 60°. This makes a new point, D, where AG intersects FE. After that, we draw a line HC so that it is parallel to FE and intersects AG at point C. This makes ΔABC ~ ΔADE ~ ΔHAG. The challenge is to find ∠FEG if ∠HAG measures 90°.

Solution

Captivating Circles 6

This is the final Captivating Circles challenge, this challenge requires the concepts of geometry and trigonometry.

Captivating Circle #6

We begin this challenge with 2 concentric circles. The smaller circle divides the larger 1 into 2 equal pieces. We then draw 2 parallel chords in the large circle, which divide it evenly into three. They do not divide the small circle into three equal pieces. The challenge is then to find the area of the biggest piece in the smallest circle if the large circle has radius 1.

Solution